Problem: Let $R$ be the region to the right of the line $y=x+1$ and enclosed by it, the curve $y=\text{cos}(x)$, and the $x$ -axis. $y$ $x$ ${y=\text{cos}(x)}$ ${y=x+1}$ $x=-1}$ $(-1,0)$ $(0,1)$ $\left(\dfrac\pi 2,0\right)$ $ R$ A solid is generated by rotating $R$ about the line $x=-1$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int_0^1 \left[(\text{arccos}(y)+1)^2-y^2\right]dy$ (Choice B) B $\pi\int_0^1 \left[(\text{cos}(y)+1)^2-(y+2)^2\right]dy$ (Choice C) C $\pi\int_{-1}^{\frac\pi2} \left[(\text{arccos}(y)+1)^2-y^2\right]dy$ (Choice D) D $\pi\int_{-1}^{\frac\pi2} \left[(\text{cos}(y)+1)^2-(y+2)^2\right]dy$
Solution: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=\text{cos}(x)}$ ${y=x+1}$ $x=-1}$ $(-1,0)$ $(0,1)$ $\left(\dfrac\pi 2,0\right)$ Let the thickness of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ This is called the washer method. What we now need is to figure out the expressions of $r_1(y)$ and $r_2(y)$, and the interval of integration. $r_1(y)$ is equal to the distance between the curve $y=\text{cos}(x)$ and the line $x=-1$. To find it, we need to solve the equation for $x$ : $x=\text{arccos}(y)$ So, ${r_1(y)=\text{arccos}(y)+1}$. $r_2(y)$ is equal to the distance between the line $y=x+1$ and the line $x=-1$. To find it, we need to solve the equation for $x$ : $x=y-1$ So, ${r_2(y)=y}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [({\text{arccos}(y)+1})^2-({y})^2] \\\\ &= \pi [(\text{arccos}(y)+1)^2-y^2] \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=1$. So the interval of integration is $[0,1]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^1 \pi \left[(\text{arccos}(y)+1)^2-y^2\right] dy \\\\ &=\pi\int_0^1 \left[(\text{arccos}(y)+1)^2-y^2\right]dy \end{aligned}$